Question

The value of the integral ${\int }_0^{\infty }\frac{1}{1+x^4}dx$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{\sqrt{2}}$
• C. $\frac{\pi }{2\sqrt{2}}$
• D. $\pi /4$

Answer 1

The correct option is C $\frac{\pi }{2\sqrt{2}}$

$I={\int }_0^{\infty }\frac{1}{x^4+1}dx={\int }_0^{\infty }(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)})dx=\frac{1}{4}\int \frac{\sqrt{2}x+2}{x^2+\sqrt{2}x+1}dx+\frac{1}{4}\int \frac{\sqrt{2}x-2}{-x^2+\sqrt{2}x-1}dx=\frac{1}{4\sqrt{2}}\int \frac{\sqrt{2}+2x}{x^2+\sqrt{2}x+1}dx+\frac{1}{4}\int \frac{1}{x^2+\sqrt{2}x+1}dx+\frac{1}{4}\int \frac{\sqrt{2}x-2}{-x^2+\sqrt{2}x-1}dx$
Let $I_1=\frac{1}{4\sqrt{2}}\int \frac{\sqrt{2}+2x}{x^2+\sqrt{2}x+1}dx$
Substituting $t=x^2+\sqrt{2}x+1\Rightarrow dt=(2x+\sqrt{2})dx$, we get
$I_1=\frac{1}{4\sqrt{2}}\int \frac{1}{t}dt=\frac{\log t}{4\sqrt{2}}=\frac{\log (x^2+\sqrt{2}x+1)}{4\sqrt{2}}$
Now let $I_2=\frac{1}{4}\int \frac{1}{x^2+\sqrt{2}x+1}dx$
Substituting $t=x+\frac{1}{\sqrt{2}}\Rightarrow dt=dx$, we get
$I_2=\frac{1}{4}\int \frac{1}{t^2+\frac{1}{2}}dt=\frac{1}{4}\int \frac{2}{2t^2+1}=ta{n}^{-1}\left(\frac{\sqrt{2}x+1}{2\sqrt{2}}\right)$
And let $I_3=\frac{1}{4}\int \frac{\sqrt{2}x-2}{-x^2+\sqrt{2}x-1}dx$
$=\frac{1}{4}\int (\frac{\sqrt{2}x-2}{\sqrt{2}(-x^2+\sqrt{2}x-1)}-\frac{1}{-x^2+\sqrt{2}x-1})dx=-\frac{log(-x^2+\sqrt{2}x-1)}{4\sqrt{2}}-\frac{tan(1-\sqrt{2}x)}{2\sqrt{2}}$
Therefore resubstituting ${I=I}_1+I_2+I_3$
${\int }_0^{\infty }\frac{1}{x^4+1}dx={[\frac{\log (x^2+\sqrt{2}x+1)}{4\sqrt{2}}+ta{n}^{-1}\left(\frac{\sqrt{2}x+1}{2\sqrt{2}}\right)-\frac{log(-x^2+\sqrt{2}x-1)}{4\sqrt{2}}-\frac{tan(1-\sqrt{2}x)}{2\sqrt{2}}]}_0^{\infty }=\frac{\pi }{2\sqrt{2}}$